The Dirac Delta Function Potential

From FSUPhysicsWiki

Jump to: navigation, search
Quantum Mechanics A
Schrödinger Equation
The most fundamental equation of quantum mechanics; given a Hamiltonian \mathcal{H}, it describes how a state |\Psi\rangle evolves in time.
Basic Concepts and Theory of Motion
UV Catastrophe (Black-Body Radiation)
Photoelectric Effect
Stability of Matter
Double Slit Experiment
Stern-Gerlach Experiment
The Principle of Complementarity
The Correspondence Principle
The Philosophy of Quantum Theory
Brief Derivation of Schrödinger Equation
Relation Between the Wave Function and Probability Density
Stationary States
Heisenberg Uncertainty Principle
Some Consequences of the Uncertainty Principle
Linear Vector Spaces and Operators
Commutation Relations and Simultaneous Eigenvalues
The Schrödinger Equation in Dirac Notation
Transformations of Operators and Symmetry
Time Evolution of Expectation Values and Ehrenfest's Theorem
One-Dimensional Bound States
Oscillation Theorem
The Dirac Delta Function Potential
Scattering States, Transmission and Reflection
Motion in a Periodic Potential
Summary of One-Dimensional Systems
Harmonic Oscillator Spectrum and Eigenstates
Analytical Method for Solving the Simple Harmonic Oscillator
Coherent States
Charged Particles in an Electromagnetic Field
WKB Approximation
The Heisenberg Picture: Equations of Motion for Operators
The Interaction Picture
The Virial Theorem
Commutation Relations
Angular Momentum as a Generator of Rotations in 3D
Spherical Coordinates
Eigenvalue Quantization
Orbital Angular Momentum Eigenfunctions
General Formalism
Free Particle in Spherical Coordinates
Spherical Well
Isotropic Harmonic Oscillator
Hydrogen Atom
WKB in Spherical Coordinates
Feynman Path Integrals
The Free-Particle Propagator
Propagator for the Harmonic Oscillator
Differential Cross Section and the Green's Function Formulation of Scattering
Central Potential Scattering and Phase Shifts
Coulomb Potential Scattering

A delta function potential, V(x)=V_0\delta(x-a)\!, is a special case of the symmetric finite square well; it is the limit in which the width of the well goes to zero and the depth of the well goes to infinity, while the product of the height and depth remains constant. For such a potential, the wave function is still continuous across the potential, ie. x=a\!. However, the first derivative of the wave function is not.

For a particle subject to an attractive delta function potential, V(x)=-V_0\delta(x),\! the Schrödinger equation is

-\frac{\hbar^2}{2m}\frac{d^2 \psi(x)}{dx^2}-V_0\delta(x)\psi(x)=E\psi(x).

For  x \neq \!0 the potential term vanishes, and all that is left is

\frac{d^2\psi(x)}{dx^2}+\frac{2mE}{\hbar^2}\psi(x)=0.

One or more bound states may exist, for which E<0\!, and  \psi(x) \! vanishes at  x = \pm \infty \!. The bound state solutions are given by

 \psi_{I}(x) = Ae^{kx},\, x < 0  \!
 \psi_{II}(x) = Be^{-kx},\, x > 0 \!

where k=\frac{\sqrt{-2mE}}{\hbar}.

The first boundary condition, the continuity of  \psi(x) \! at  x = 0 \!, yields  A = B \! .

The second boundary condition, the discontinuity of  \frac{d\psi(x)}{dx} \! at  x = 0 \!, can be obtained by integrating the Schrödinger equation from  -\epsilon \! to \epsilon \! and then letting  \epsilon \rightarrow 0 \!

Integrating the whole equation across the potential gives

-\frac{\hbar^2}{2m}\int_{-\epsilon}^{\epsilon} \frac{d^2 \psi(x)}{dx^2}\,dx-V_0\int_{-\epsilon}^{+\epsilon} \delta(x)\psi(x)\,dx=\int_{-\epsilon}^{\epsilon} E \psi(x)\,dx

In the limit \epsilon \rightarrow 0 \!, we have

-\frac{\hbar^2}{2m}\left[\frac{d \psi_{II}(0)}{dx}-\frac{d \psi_{I}(0)}{dx}\right]-V_0\psi(0)=0,

which yields the relation,  k = \frac{mV_0}{\hbar^2}. \!.

Since we defined  k = \frac{\sqrt{-2mE}}{\hbar}, \!, we have  \frac{\sqrt{-2mE}}{\hbar} = \frac{mV_0}{\hbar^2} \!. Then, the energy is  E = -\frac{mV_0^2}{2\hbar^2} \!

Finally, we normalize  \psi(x) \!:

\int_{-\infty}^{\infty} |\psi(x)|^2\,dx=2|B|^2\int_{0}^{\infty} e^{-2kx}\,dx=\frac{|B|^2}{k}=1

so,

B=\sqrt{k}=\frac{\sqrt{mV_{0}}}{\hbar}

Evidently, the delta function well, regardless of its "strength"  V_{0} \!, has one bound state:

 \psi(x)=\frac{\sqrt{mV_{0}}}{\hbar}e^{\frac{-mV_{0}|x|}{\hbar^{2}}}

For a delta potential of the form V_0\delta(x-a)\!, we may apply a similar procedure to the one employed above to show that the discontinuity in the derivative of the wave function is

\frac{d \psi(a^+)}{dx}-\frac{d \psi(a^-)}{dx}=\frac{2m V_0}{\hbar^2}\psi(a).

Problem

(Double delta function potential)

Consider a double delta function potential, V(x)=-g\delta(x-a)-g\delta(x+a).\! Prove that a ground state with even parity always exists. By means of a sketch, show that a first excited state also exists for a sufficiently large potential strength g\! and sufficiently large well separation 2a.\!

Solution

External Link

Additional information on the Dirac delta function

Personal tools